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sergejj [24]
3 years ago
11

Calculate the energy required to freeze 378 grams of water to ice at 0°c​

Chemistry
2 answers:
ioda3 years ago
7 0

Answer:

What was the answer?

Explanation:

I have the same question

DerKrebs [107]3 years ago
3 0

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In a pendulum, a continuous change occurs between kinetic energy and
Masja [62]
It should be potential energy!!!
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earnstyle [38]

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I think B or D

Explanation:

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4 0
3 years ago
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How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O)? For example if th
oee [108]

Given :

Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .

To Find :

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .

Solution :

By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

So , volume of solution does not matter .

Moles of oxygen , n=\dfrac{2.666}{16}=0.167\ mole .

Now , molecule of CO contains 1 mole of C .

So , moles of C is also 0.167 mole .

Mass of carbon , m=12\times 0.167=2\ g .

Therefore , mass of carbon is 2 grams .

Hence , this is the required solution .

5 0
3 years ago
What product of an acid base reaction is an ionic compound apex?
charle [14.2K]
When acids react with bases they produce salt and water such as:
HCl + NaOH → NaCl + H₂O
According to strength of acid and base, we have 4 types of salts:
salt of strong acid and strong base like: NaCl
salt of weak acid and strong base like: CH₃COONa
salt of strong acid and weak base like: NH₄Cl
salt of weak acid and weak base like: CH₃COONH₄
8 0
3 years ago
Read 2 more answers
A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m
Lana71 [14]

Answer:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol

Moles of hydrogen gas produced = 0.01225 mol

2M+2xHCl\rightarrow 2MCl_x+xH_2

Moles of metal =\frac{0.225 g}{27.0 g/mol}=8.3333 mol

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

\frac{8.3333}{0.01225 mol}=\frac{2}{x}

x = 2.9 ≈ 3

2M+6HCl\rightarrow 2MCl_3+3H_2

MCl_3\rightarrow M^{3+}+Cl^-

Formulas for the oxide and sulfate of M will be:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3.

3 0
3 years ago
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